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(a) Let $X$ and $Y$ be two independent discrete random variables. Derive a formula for expressing the distribution of the sum $S = X + Y$ in terms of the distributions of $X$ and of $Y$.

(b) Use your formula in part (a) to compute the distribution of $S = X +Y$ if $X$ and $Y$ are both discrete and uniformly distributed on $\{1,\dots,K\}.$

(c) Suppose now $X$ and $Y$ are continuous random variables with densities $f$ and $g$ respectively ($X,Y$ still independent). Based on part (a) and your understanding of continuous random variables, give an educated guess for the formula of the density of $S = X +Y$ in terms of $f$ and $g$.

(d) Use your formula in part (c) to compute the density of $S$ if $X$ and $Y$ have both uniform densities on $[0, a]$.

(e) Show that if $X$ and $Y$ are independent normally distributed variables, then $X +Y$ is also a normally distributed variable.

Added from comment: What I think I have figured out so far is:

a) distribution of $S$ is $$\text{mean}(S) = \text{mean}(X) + \text{mean}(Y)$$ and $$\text{stddev}(S) = \sqrt{\text{stddev}(X)^2 + \text{stddev}(Y)^2}$$

b) stuck a little on b... help please

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What I think I have figured out so far is: a) distribution of S is Smean = mean of X + mean of Y and Sstanddev = sqrt(stddevX^2 + stddevY^2) b) stuck a little on b... help please – Herman Chong Nov 17 '11 at 9:12
For a), use $P[X+Y=k]=\sum_i P[X=k-i , Y=i] = \sum_i P[X=k-i]P[Y=i]$. For b), you know the pmf's of $X$ and $Y$, so just substitute into the formula from a). – David Mitra Nov 17 '11 at 9:18
hmm...i'm confused as to where you got that equation/expression from and also are for part b) is pmf is 1/k because of uniform distrib – Herman Chong Nov 17 '11 at 9:23

In a), you are asked to find the distribution of $S$, not the mean and variance. The distribution of $S$ is by definition the function $p$ that gives point probabilities. You need to find the rule for this function $$p(k)=P[X+Y=k].$$

For simplicity, I'll assume $X$ and $Y$ take nonnegative integer values. Now, fix a value of $k$. Then $X+Y=k$ if and only if one of the following mutually exclusive alternatives hold: $$ X=k,Y=0;\quad X=k-1,Y=1;\quad X=k-2,Y=2;\quad \cdots; X=0,Y=k. $$ From this, it follows that $$ p(k) =\sum_{i=0}^k P[X=k-i, Y=i]. $$ Since $X$ and $Y$ are independent, $P[X=k-i, Y=i]=P[X=k-i]P[Y=i]$. Thus $$ p(k) =\sum_{i=0}^k P[X=k-i]P[Y=i]=\sum_{i=0}^k p_X(k-i)p_Y(i), $$ where $p_X$ is the distribution of $X$ and $p_Y$ is the distribution of $Y$.

In general, the distribution of $X+Y$ is the convolution of the distributions of $X$ and $Y$. If $X+Y$ takes the value $z$: $$ p_{X+Y}(z) =\sum_{y} p_X(z-y)p_Y(y), $$
where the sum is taken over all values of $Y$.

For part b), use the particular distributions for $X$ and $Y$ and substitute into the above formula.

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